Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 23, Problem 69AP

(a)

To determine

The electric charge at (2.00m,0)

(a)

Expert Solution
Check Mark

Answer to Problem 69AP

The electric field at (2.00m,0) is 24.2i^N/C.

Explanation of Solution

Conclusion:

Write the expression for the electric field.

    E=(E1+E2+E3)i^                                                                                                   (I)

Here, s is the total electric field, E1 is the field about the charge 4.00nC, E2 is the field about the charge 5.00nC, E3 is the field about the charge 3.00nC.

Write the expression for the electric field.

    E=kqr2                                                                                                                     (II)

Here, k is the constant, q is the electric charge, r is the distance from the position considered.

Substitute kqr2 in equation (I) and rewrite the equation.

    E=(kq1r12+kq2r22+kq3r32)i^                                                                                         (III)

The following diagram shows the distance of the charges from (2.00m,0).

Physics for Scientists and Engineers With Modern Physics, Chapter 23, Problem 69AP , additional homework tip  1

Figure-(1)

Substitute 9×109Nm2/C2 for k, 4.00nC for q1, 2.500m r1, 5.00nC for q2, 2.00m for r2, 3.00nC for q3, 1.200m for r3 in the equation (III) to get E.

    E=(9×109Nm2/C2[4.00nC(2.500m)2+5.00nC(2.00m)2+3.00nC(1.200m)2])i^=(9×109Nm2/C2[0.64nC/m2+1.25nC/m2+2.083nC/m2])i^=(9×109Nm2/C2×2.693×109C/m2)i^=24.2i^N/C

Therefore, the electric field at (2.00m,0) is 24.2i^N/C.

(b)

To determine

The electric field at (0,2.00m).

(b)

Expert Solution
Check Mark

Answer to Problem 69AP

The resultant electric field is 9.4 N/C and the direction is 63.4° above xaxis.

Explanation of Solution

Rewrite the equation (III) for resolving the forces along x direction.

    Ex=[kq1r12cosθ1+kq2r22cosθ2kq3r32cosθ3]i^                                                         (IV)

Rewrite the equation (III) for resolving the forces along y direction.

    Ey=[kq1r12sinθ1+kq2r22sinθ2+kq3r32sinθ3]j                                                              (V)

Write the expression to calculate the electric field at (0,2.00m).

    E=Ex+Ey                                                                                                            (VI)

Write the expression for the resultant electric field.

    ER=(Ex)2+(Ey)2                                                                                           (VII)

Here, ER is the resultant electric field.

Write the expression to calculate the direction.

  θ=tan1(EyEx)                                                                                                    (VIII)

Here, θ is the angle.

Conclusion:

The following diagram shows the distance of the charges from (0,2.00m).

Physics for Scientists and Engineers With Modern Physics, Chapter 23, Problem 69AP , additional homework tip  2

Figure-(2)

From the figure-(2) calculate the distance of the charge from the position (0,2.00m).

Consider the triangle ABO,

    tanθ1=2.00m0.500mθ1=tan1(4)θ1=75.96°

Consider the triangle ABO,

    sinθ1=2.00mr1r1=2.00msin75.964°=2.06m

Consider the triangle CBO,

    tanθ3=2.00m0.800mθ3=tan1(2.5)θ3=68.19°

Consider the triangle CBO,

    sinθ3=2.00mr3r3=2.00msin68.19°=2.15m

Substitute 9×109Nm2/C2 for k, 4.00nC for q1, 2.06m r1, 75.96° for θ1, 5.00nC for q2, 2.00m for r2, 90° for θ2, 3.00nC for q3, 2.15m for r3, 68.19° for θ3 in the equation (III) to get E.

    Ex=[9×109Nm2/C2[4.00nC(2.06m)2(cos75.96°)+5.00nC(2.00m)2(cos90°)3.00nC(2.15m)2(cos68.19°)]]i^=(9×109Nm2/C2[0.228nC/m2+0nC/m2+0.241nC/m2])i^=(9×109Nm2/C2×(0.469nC/m2×109CnC))i^=4.21i^N/C

Substitute 9×109Nm2/C2 for k, 4.00nC for q1, 2.06m r1, 75.96° for θ1, 5.00nC for q2, 2.00m for r2, 90° for θ2, 3.00nC for q3, 2.15m for r3, 68.19° for θ3 in the equation (III) to get E.

    Ey=[9×109Nm2/C2[4.00nC(2.06m)2(sin75.96°)+5.00nC(2.00m)2(sin90°)+3.00nC(2.15m)2(sin68.19°)]]j^=(9×109Nm2/C2[0.914nC/m2+1.25nC/m2+0.60nC/m2])j^=(9×109Nm2/C2×(0.842nC/m2××109CnC))j^=8.42j^N/C

Substitute 4.21i^N/C for Ex and 8.42j^N/C for Ey in equation (VI) to solve for E.

    E=4.21i^N/C+8.42j^N/C=(4.21i^+8.42j^)N/C

Then the resultant force is,

Substitute for 4.21i^N/C for Ex and Ey 8.42j^N/C in  (VII).

    ER=(4.21i^N/C)2+(8.42j^N/C)2=9.4 N/C

Substitute for 4.21i^N/C for Ex and Ey 8.42j^N/C in  (VIII).

Direction of the resultant force is,

    θ=tan1(8.42j^N/C4.21i^N/C)=63.4°

Therefore, resultant electric field is 9.4 N/C and the direction is 63.4° above xaxis.

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Chapter 23 Solutions

Physics for Scientists and Engineers With Modern Physics

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